Question

The enthalpy of combustion of \(\mathrm{CH}_{4}(g)\) when \(\mathrm{H}_{2} \mathrm{O}(l)\) is formed is \(-891 \mathrm{~kJ} / \mathrm{mol}\) and the enthalpy of combustion of \(\mathrm{CH}_{4}(\mathrm{~g})\) when \(\mathrm{H}_{2} \mathrm{O}(g)\) is formed is \(-803 \mathrm{~kJ} / \mathrm{mol}\). Use these data and Hess's law to determine the enthalpy of vaporization for water.

Short answer

The enthalpy of vaporization for water is \(88 \, kJ/mol\).

Step-by-step solution

Write the chemical equations for combustion of methane with different products

Equation 1: Combustion of methane with the formation of liquid water: \[CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l)\] Enthalpy change for Equation 1: \(-891 \, kJ/mol\) Equation 2: Combustion of methane with the formation of gaseous water: \[CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(g)\] Enthalpy change for Equation 2: \(-803 \, kJ/mol\)

Get the chemical equation for enthalpy of vaporization

From the given data, we want to find the enthalpy of vaporization, which represents the conversion of liquid water to gaseous water: Equation 3: Enthalpy of vaporization of water: \[2H_{2}O(l) \rightarrow 2H_{2}O(g)\] By Hess's law, we can manipulate and add the given equations to find the enthalpy of vaporization:

Use Hess's law to solve for the enthalpy of vaporization

Our goal is to manipulate Equations 1 and 2 to isolate Equation 3. To do this, we can subtract Equation 1 from Equation 2: \[(-803 \, kJ/mol) - (-891 \, kJ/mol) = \Delta H\] where \(\Delta H\) is the enthalpy change for Equation 3. Calculate the enthalpy of vaporization: \[\Delta H = -803 \, kJ/mol + 891 \, kJ/mol\] \[\Delta H = 88 \, kJ/mol\] Our final result is:

Enthalpy of vaporization for water

The enthalpy of vaporization for water is \(88 \, kJ/mol\).

Key concepts

Enthalpy of Vaporization

When we talk about the enthalpy of vaporization, we're referring to the amount of energy needed to turn a liquid into a gas at a constant temperature and pressure.
For water, this process can be represented by the equation:

• \[2H_{2}O(l) \rightarrow 2H_{2}O(g)\]

The enthalpy change associated with this transformation is the enthalpy of vaporization. It's a crucial concept in chemistry, as it explains how heat energy is absorbed to overcome the forces that hold the water molecules together in a liquid state.
The energy input breaks these intermolecular forces, allowing the water molecules to escape and become vapor.
In this exercise, the enthalpy of vaporization of water is derived from the combustion data of methane, demonstrating how different chemical reactions are intricately connected.

Hess's Law

Hess's law is a principle that states the total enthalpy change for a chemical reaction is the same, regardless of the pathway taken.
This concept can simplify complex chemical calculations by using known enthalpy changes of related reactions to find unknown values.
In the exercise, we used Hess's law to determine the enthalpy of vaporization for water.
This was achieved by manipulating the enthalpy changes from the combustion of methane in different states of water.
By writing two combustion equations with initial and final states and then subtracting them, we could discover the enthalpy required for the vaporization process:

• Equation 1: Formation of water as a liquid
• Equation 2: Formation of water as a gas

Finally, by subtracting the enthalpy of Equation 1 from Equation 2, we determined the enthalpy of vaporization for water as 88 kJ/mol. So, even though Hess's law might sound complex, it merely allows us to add up small steps to find the bigger picture!

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show the reactants transforming into products and are balanced to ensure matter conservation.
In our exercise, we dealt with combustion reactions and the transition from liquid to gaseous water.

• Reactants and Products: The chemical equations displayed the conversion of methane and oxygen into carbon dioxide and water, either in liquid or gaseous form.
• Balancing: Each side of the equation must have the same number of atoms of each element, which is achieved through stoichiometry.

In the exercise, we had:

• Combustion of methane with liquid water formation: \[CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l)\]
• Combustion of methane with gaseous water formation: \[CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(g)\]

These equations represented the transformations and provided us the pathways that led to deriving the enthalpy of vaporization.
Understanding and writing chemical equations is essential, as they are the language of chemistry that conveys how substances interact and transform in reactions.